My friend right now I'm a student and I'm broke af, but the moment I graduate ad I get a job I'm fucking going to hit that 200$ donation button, thank you for the immense help you are giving us, *ak lectures is the best!!* a big hug from Italy :)
Saved my day again, by the way thank you so much literally, all the students in my university watch your lectures and I got all my friends to watch your lectures too , keep up the good work ❤️
if i was taught like this from begining i would be a very very intelligent student. but sorry to say that our teachers have a lot knowledge but they have not the potential to teach like this. thank you sir . love from india.
Hi, thanks for this lecture about Thermodynamics of chemical reactions. I have a small concern with the upper right part of your board; It seems that you forgot the standard gibbs free energy (DelatG°). In fact "DeltaG = DeltaG° + RTLnKeq", and when reaching the equilibrium, DeltaG equals zero means that DeltaG° = -RTlnKeq (but does not mean that k1=k2). Otherwise all equilibrium constants would be equal to 1. In other words reaching the equilibrium allows one to quantify Keq, the equilibrium constant (which is also : Keq=k1/k2) and this value not necessarily equals to one. I would rather say that reaching the equilibrium means that k1*[substrate]=k2*[product] (forward velocity equals reverse velocity) The case you are mentioning, (i.e. when k1=k2) is just a particular case where reactants and products have the same level of energy and they are both at the same concentration at equilibrium (50% of reactants + 50% of products at equilibrium). Or that "k1=k2" means DeltaG = DeltaG° and DeltaG is independent of the temperature at wich the reaction takes place. But I agree with the overall message that at equilibrium the kinetic of the reaction has stopped and DeltaG equals zero.
Please help me understand this. I was able to follow until the 11th minute where you say that the enzyme lowers the activation energy hence allows the reaction to reach equilibrium faster (this I understand) but now what confuses me is that you said that lowering of the activation energy results in stabilising the transition state. Now, won't stabilising the transition state mean that it will take longer for the equilibrium to be reached because the components of the transition state are no longer unstable - isn't it that instability is associated with readiness to react and stability takes a longer time to reach?
I think is late to answer this question but in principle you asked a very good question. What I understand about catalyst or rather enzymes is that they lower the barrier which is the hill and this allow more molecules to overcome it to form product at a faster rate. The term stabilization has to do with lowering of activation energy but in principle the molecules on the top of the hill are still not thermodynamically stable and eventually will react to form product with lower Gibbs free energy and now the driving force is the reduction of Gibbs free energy. Note, that catalyst only remove or reduce the barrier but don't determine which product is to be formed. That has to be determined by thermodynamics. So enzymes or catalysts only affect kinetics but not thermodynamics. So to answer your question, it will not stay at the hill for longer time due to its instability, it will further decompose to be at the lowest energy state i.e. thermodynamic stable state
Thanks for explaning....You do a great job....i had a question regarding activation energy of surface chemistry... amongst physisorption and chemisorption,which of the two has greater activation energy?
+Akshat Bajpai It is evidently Chemisorption , simply because it involves a chemical reaction and needs activation energy . On the other hand , there is NO reaction taking place between the adsorbant and adsorbate during physisorption (but physical adsorption of gases involve condesation ). Am I clear ?
Hello, I know this isn't directly related to the video but you have talked about this before in another video so I'd like to ask- does chymotrypsin also cleave at the c terminus of methionine? Or does it only cleave after large aromatic residues? Also, I read somewhere that similar to typsin, chymotrypsin only cleaves at the c terminus of large hydrophobic residues ** UNLESS THE next amino acid is proline- is this true?
